Answer
$$\frac{{dy}}{{d\theta }} = \frac{{\theta \sin \theta }}{{\sqrt {\sec \theta } }}\left( {\frac{1}{\theta } + \cot \theta - \frac{1}{2}\tan \theta } \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{{\theta \sin \theta }}{{\sqrt {\sec \theta } }} \cr
& {\text{Take the natural logarithm of both sides:}} \cr
& {\text{Use the properties of logarithms}} \cr
& \ln y = \ln \left( {\frac{{\theta \sin \theta }}{{\sqrt {\sec \theta } }}} \right) \cr
& {\text{quotient rule:}} \cr
& \ln y = \ln \left( {\theta \sin \theta } \right) - \ln \left( {\sqrt {\sec \theta } } \right) \cr
& {\text{product rule:}} \cr
& \ln y = \ln \left( \theta \right) + \ln \left( {\sin \theta } \right) - \ln \left( {\sqrt {\sec \theta } } \right) \cr
& {\text{power rule:}} \cr
& \ln y = \ln \left( \theta \right) + \ln \left( {\sin \theta } \right) - \frac{1}{2}\ln \left( {\sec \theta } \right) \cr
& {\text{Take derivatives of both sides with respect to }}\theta \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{1}{\theta } + \frac{{\cos \theta }}{{\sin \theta }} - \frac{1}{2}\left( {\frac{{\sec \theta \tan \theta }}{{\sec \theta }}} \right) \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{1}{\theta } + \cot \theta - \frac{1}{2}\tan \theta \cr
& {\text{solve for }}\frac{{dy}}{{d\theta }} \cr
& \frac{{dy}}{{d\theta }} = y\left( {\frac{1}{\theta } + \cot \theta - \frac{1}{2}\tan \theta } \right) \cr
& {\text{substitute }}\frac{{\theta \sin \theta }}{{\sqrt {\sec \theta } }}\sin \theta {\text{ for }}y{\text{ }} \cr
& \frac{{dy}}{{d\theta }} = \frac{{\theta \sin \theta }}{{\sqrt {\sec \theta } }}\left( {\frac{1}{\theta } + \cot \theta - \frac{1}{2}\tan \theta } \right) \cr} $$
